Question

A sample of 250 people resulted in a confidence interval estimate of 0.14 and 0.20 for the proportion of people who prefer sliding glass doors to traditional back doors. Based on this information, what was the confidence level used in this estimation?

There is not enough information provided to determine the answer.

Almost 0.40

About 0.95

Approximately 0.79

Answer 1

Solution Given, is n = 250 confidence interval for proportion p (0.14 0.20) je. Upper limit 20.20 Toner limit =0.14 Formula for confidence interval for p is IZ JALID Die B I mangin of error Upper limit = R + margin of error = P + Zes B(1-P) Lower limit = B - Margin of error ? - Za BCI-B) from D & D we can say margin of error Upper limit - lower limit = 0.20 -0.14 margin of emor = 0.03. and Upper limet + lower limbo 0-20 + 0.14 = 0.17 SO

Now consider margin or error = P2 - 0.03 - 24 / 171 - 0 -17) 250 za 2 0.03 017(1-0.17) 250 ..za - 1.26 use table. See the colomo of 1.2 in this colomon & Han see value in the row at .06. This value is I-& 2. See corresponding ::.- t = 0.896 1 - 0.896 - 0.104 1. -2 0.104 = 0.208 Now, confidence level =1-X = 1-0.208=0.792 confidence lever is 0.792 je. 79.2% Answer is approximately 0.79